Question 828326
ONE WAY TO DO IT:
{{{14-1&1/2=13&2/2-1&1/2=12^1/2}}}
writing {{{14}}} as {{{14=13+1=13+2/2=13&2/2}}}
and subtracting
{{{13&2/2-1&1/2=(13+2/2)-(1+1/2)=13+2/2-1-1/2=(13-1)+(2/2-1/2)=12+1/2=12&1/2}}}
It is what you would do if you had ${{{14}}} and had to pay ${{{1&1/2}}} with exact change.
You would ask for change for $1, and get two ${{{1/2}}} ($0.50) coins in its place.
You would have $13 in bills that you would use to pay the $1 part of ${{{1&1/2}}} ,
keeping the other $13-$1=$12 in bills,
and you would use 1 of the 2 ${{{1/2}}} coins to pay the rest of the ${{{1&1/2}}} ,
keeping the other ${{{1/2}}} coin
That would leave $12 in bills and ${{{1/2}}} in coin in your wallet.
 
ANOTHER WAY
{{{14-1&1/2=14-3/2=28/2-3/2=(28-3)/2=25/2=24/2+1/2=12+1/2=12&1/2}}}